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DiKsa [7]
3 years ago
11

If the resistance is not changed but the voltage is increased, what happens to the current

Physics
1 answer:
julsineya [31]3 years ago
6 0

We know that According to Ohm's Law :

Current passing through a Conductor is directly proportional to the Voltage over a given Resistance.

⇒ V ∝ I

⇒ V = I × R

If Resistance is not changed and Voltage is increased, Based on Ohm's law we can conclude that Current flowing will also increase, because Voltage is directly proportional to Current.

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<h2>Answer: </h2><h2>- Jupiter has orbiting moons.</h2><h2>- The Sun has sunspots and rotates on its axis.</h2><h2>- The Moon has mountains, valleys, and craters.</h2><h2>- Venus goes through a full set of phases.</h2>

Explanation:

In 1609 Galileo built a telescope, with which he observed mountains and craters on the Moon, discovered Jupiter’s major satellites and the next year he published these discoveries in his book <em>The Sidereal Messenger</em>.

In addition, Galileo observed that Venus presented phases (such as those of the moon) together with a variation in size; observations that are only compatible with the fact that Venus rotates around the Sun and not around Earth. This is because <u>Venus presented its smaller size when it was in full phase and the largest size when it was in the new one, when it is between the Sun and the Earth.  </u>

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On the other hand, <u>although Galileo was not the first to observe sunspots</u>, he gave the correct explanation of their existence, which supported the idea that planets revolve around the Sun.

These observations and discoveries were presented by Galileo to the Catholic Church (which supported the geocentric theory at that time) as a proof that completely refuted Ptolemy's geocentric system and affirmed Copernicus' heliocentric theory.

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3 years ago
BRAINLIEST+10PTS: does ozone absorb just harmful UV rays? Or just a random portion of UV?
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Answer:As sunlight passes through the atmosphere, all UVC and most UVB is absorbed by ozone, water vapour, oxygen and carbon dioxide. UVA is not filtered as significantly by the atmosphere. Is there a connection between ozone depletion and UV radiation? Ozone is a particularly effective absorber of UV radiation.

Explanation:

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2 years ago
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A student has a thin copper beaker containing 100 g of a pure metal in the solid state. The metal is at 215°C, its exact melting
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Answer:

The metal will melt but their will be no change in temperature.

Explanation:

The metal is at its melting temperature which means it is still in solid phase but have to cross the enthalpy of its condensation at this same temperature to convert into liquid phase.

<u>On supplying heat, the metal's temperature will not change as the heat will be required as enthalpy of condensation to melt the solid to liquid at the melting temperature.</u>

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3 years ago
Name and describe two forces that act on your body as you walk down the hallway.
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3 years ago
A bare helium nucleus has two positive charges and a mass of 6.64×10−27kg(a) Calculate its kinetic energy in joules at 2.00% of
grandymaker [24]

To solve this problem we will apply the concepts related to kinetic energy and energy conservation. The kinetic energy will be expressed in terms of mass and speed, as well as load and voltage. From this last expression we will find the charges by electron and by Helium nucleus.

a ) Kinetic Energy is given as

KE = \frac{1}{2} mv^2

Replacing with our values we have that

KE = \frac{1}{2} (6.64*10^{-27})(2.0\% (3.00*10^8))^2

KE = 1.1935*10^{-13}J

Therefore the kinetic energy of the helium nucleus is 1.1935*10^{-13}J

PART B) Now for calculate the electron volts we use the kinetic energy as a expression between the charge and the voltage, that is

KE = qV

Here,

q = Charge of an electron

V = Voltage

Rearranging to find the potential we have,

V = \frac{KE}{q}

V = \frac{1.19*10^{-13}}{1.6*10^{-19}}

V = 743750eV

Therefore the kinetic energy in electron vols is 743750eV

PART C) Applying the same relationship but now using the Helium core load, we will have to

KE = QV

Here,

Q = Charge of a helium nucleus

V = Voltage

Rearranging to find the potential we have

V = \frac{KE}{Q}

But we need to note that the charge is equal to the number of charge for the unit charge, then

Q = \text{No. Charge} \times \text{Unit Charge}

Q = (2)(1.6*10^{-19}C)

Q = 3.2*10^{-19}C

Now replacing we have that

V= \frac{1.19*10^{-13}}{32*10^{-19}}

V = 371875V

Therefore the voltage applied is  371875V

5 0
3 years ago
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